Найдите корень уравнения 0,5^6-2x = 32

Question

Алгебра: Найдите корень уравнения 0,5^6-2x = 32

Zelka20199 · Accepted Answer

X = π/3 - y
sin(π/3 - y)*siny = 1/4
(sin(π/3)*cosy - siny*cos(π/3) )*siny = 1/4
((cosy)*√3/2 - (siny)/2)*siny = 1/4
(cosy*siny*√3 - sin^2(y))/2 = 1/4
√3*cosy*siny - sin^2(y) = 1/2
1/2 = 0.5sin^2(y) + 0.5cos^2(y)
√3*cosy*siny - sin^2(y) - 0.5sin^2(y) - 0.5cos^2(y) = 0 - делим на -0.5
cos^2(y) + 3sin^2(y) - 2√3*cosy*siny = 0 - делим на cos^2(y)
1 + 3tg^2(y) - 2√3*tgy = 0
замена tg(y) = t
3t^2 - 2√3*t + 1 = 0
(√3t - 1)^2 = 0
√3t = 1, t = √3/3
tg(y) = √3/3
y = π/3 + πk
x = π/3 - π/3 - πk = -πk
ответ: x = -πk, y = π/3 + πk

MOGZ ответил